# I Function of 2 variables, max/min test, D=0 and linear dependence

1. Dec 19, 2017

### binbagsss

$f(x,y)$

a critical point is given by $f_x=0$ and $f_y=0$ simultaneously.

the test is:

$D=f_{xx}f_{yy}-(f_{xy})^2$

if $D >0$ and $f_{xx} <0$ it is a max
if $D >0$ and $f_{xx} >0$ it is a min
$D >0$ is is a saddle
if $D =0$ it is inconclusive, and $f_x$ and $f_y$ are not linear independent.

I'm stuck on the $D=0$ comment re linear independence. So is this saying that $x$ and $y$ are not linear indepedent?

So if i take an arbitary function $f(x,y)$ and $y=h(x)$, h some linear function, then I should get $D=0$ or not?

2. Dec 19, 2017

### Staff: Mentor

It is saying that the two first partials are linearly dependent, not that x and y are linearly dependent.

3. Dec 19, 2017

### binbagsss

What is characteristic of a function whose partials are linearly dependent?

4. Dec 19, 2017

### Staff: Mentor

If two functions (partials in this case) are linearly dependent, then each will be a constant multiple of the other.

5. Dec 19, 2017

### binbagsss

yup, so if i have a $f(x,y)$ function all term must be a multiple of $(xy)^n$ , $n \in R$ or if it contains terms solely of $x$ or $y$ they must be of at most the power of $2$? (To conclude the partials $f_x$ and $f_y$ are linear dependent, not higher partials)?

edit: re second comment is incorrect, but would yield $D=0$ without implying that $f_x$ and $f_y$ are linearly independent? so in OP comment only holds for higher powers? or a function without sole terrms of x or y if the power is lower

6. Dec 19, 2017

### Staff: Mentor

I haven't worked out an example. You want a function f(x, y) such that fx = fy = 0, and for which the discriminant D = 0.

7. Dec 19, 2017

### binbagsss

apologies edited post since above

8. Dec 19, 2017

### StoneTemplePython

this is a typo and should say less than zero.

This is a lot easier in my view if you know bits about linear algebra. Do you know what a Hessian is? How about a quadratic form?

consider the 2 variable function (I used $x_1, x_2$ instead of $x,y$:

edit: (due to some domain subtleties, the below is a better example than the prior function I mentioned)

$f(x_1,x_2) = (x_1 + x_2)^2 = x_1^2 + x_2^2 + 2x_1 x_2$

The Hessian is given by
$\mathbf H = \begin{bmatrix} 2 & 2\\ 2 & 2 \end{bmatrix} = 2 \big(\mathbf 1 \mathbf 1^T\big)$

This is a rank one matrix and hence has determinant of zero. (Equivalently compute the determinant directly and see $2*2 - 2*2 = 0$ )

Last edited: Dec 19, 2017